A swimming pool 30m long and 10m wide has a vertical sides and bottom at slope. The depth of water at shallow and deep sides are 2m and 5m respectively. Two outlets, each 0.4m diameter have been provided at each of deep and shallow ends. Calculate time taken to empty pool if both outlets are open. Take `C_d` as 0.6 for both openings.

SOLUTION:

Here,

Area of orifice,`a=(pi**0.4^2)/4=0.1256\ m^2`

Area of pool,`A=31**10=300 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water ab....Show More

SOLUTION:

Here,

Area of orifice,`a=(pi**0.4^2)/4=0.1256\ m^2`

Area of pool,`A=31**10=300 m^2`

Let,

`T_1` be the time required to bring water level from AB to CD.

`T_2` be the time to empty the water below the portion CD

Thus the total time to empty the tank is,

`T=T_1+T_2`... (i)

Initially, let us consider an instant such that the height of water above the centre of upper orifice is h m. Let us say that during time `dT`, the level of water fall by`dh`.

Now from the principle of continuity, 

`-A*dh=(Q_1+Q_2)dT`

`-A*dh=C_d*a*root ()(2g)*(root ()(h)+root ()(h+3))`

`dT=-A/(C_d*a*root ()(2g))*(dh)/(root ()(h)+root ()(h+3))`

The total time required to drop level of water from AB to CD can be obtained by integrating this equation from limit h=2 to h=0 since h is assumed to be a  depth of water surface above upper orifice.

`T_1=-A/(C_d*a*root ()(2g))\int_{2}^{0}(dh)/(root ()(h)+root ()(h+3))`

`=-A/(C_d*a*root ()(2g))\int_{2}^{0} (root ()(h)-root ()(h+3))/(-3)dh`

Solving this using calculator,we get,

`T_1=630.3 sec`

Again, let us assume that the water is at any height h above the bottom orifice at any instant of time t within the triangular element and let the level of water falls by an amount dh in small time interval dT.

Width of water surface is,

`=h/3**10`

And the area of water surface is,

`A=(h/3**10)**30=100h`

Now from the principle of continuity, 

`-A*dh=C_d*a*root ()(2gh)*dT`

The total time required to drop level of water from CD to E (i.e, empty tank) can be obtained by integrating this equation from limit h=3 to h=0 since h is assumed to be a  depth of water surface above lower orifice i.e,bottom of tank.

`T_2=-A/(C_d*a*root ()(2g))\int_{3}^{0}1/(root ()(h))dh`

`=-100/(C_d*a*root ()(2g))\int_{3}^{0}root ()(h)dh`

Solving this using calculator, we get,

`T_2=1037.7 sec`

Thus the total time required to completely emptying the tank is,

`T=T_1+T_1`

`T=630.3+1037.7`

`T=1668 sec`